3.27.67 \(\int \frac {e^{-x} (2-4 x-20 e^x x-x \log (x^2))}{4 x} \, dx\) [2667]

Optimal. Leaf size=25 \[ -5 x+7 (-2+\log (4))+\frac {1}{4} e^{-x} \left (4+\log \left (x^2\right )\right ) \]

[Out]

1/4*(4+ln(x^2))/exp(x)+14*ln(2)-14-5*x

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Rubi [A]
time = 0.18, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 6874, 2326} \begin {gather*} \frac {e^{-x} \left (x \log \left (x^2\right )+4 x\right )}{4 x}-5 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - 4*x - 20*E^x*x - x*Log[x^2])/(4*E^x*x),x]

[Out]

-5*x + (4*x + x*Log[x^2])/(4*E^x*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {e^{-x} \left (2-4 x-20 e^x x-x \log \left (x^2\right )\right )}{x} \, dx\\ &=\frac {1}{4} \int \left (-20+\frac {e^{-x} \left (2-4 x-x \log \left (x^2\right )\right )}{x}\right ) \, dx\\ &=-5 x+\frac {1}{4} \int \frac {e^{-x} \left (2-4 x-x \log \left (x^2\right )\right )}{x} \, dx\\ &=-5 x+\frac {e^{-x} \left (4 x+x \log \left (x^2\right )\right )}{4 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.06, size = 25, normalized size = 1.00 \begin {gather*} \frac {1}{4} \left (4 e^{-x}-20 x+e^{-x} \log \left (x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 4*x - 20*E^x*x - x*Log[x^2])/(4*E^x*x),x]

[Out]

(4/E^x - 20*x + Log[x^2]/E^x)/4

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Maple [A]
time = 0.18, size = 17, normalized size = 0.68

method	result	size
default	\(-5 x +\frac {\left (4+\ln \left (x^{2}\right )\right ) {\mathrm e}^{-x}}{4}\)	\(17\)
norman	\(\left (1-5 \,{\mathrm e}^{x} x +\frac {\ln \left (x^{2}\right )}{4}\right ) {\mathrm e}^{-x}\)	\(19\)
risch	\(\frac {\ln \left (x \right ) {\mathrm e}^{-x}}{2}-\frac {\left (i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+40 \,{\mathrm e}^{x} x -8\right ) {\mathrm e}^{-x}}{8}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-x*ln(x^2)-20*exp(x)*x-4*x+2)/exp(x)/x,x,method=_RETURNVERBOSE)

[Out]

-5*x+1/4*(4+ln(x^2))/exp(x)

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Maxima [A]
time = 0.29, size = 18, normalized size = 0.72 \begin {gather*} \frac {1}{4} \, e^{\left (-x\right )} \log \left (x^{2}\right ) - 5 \, x + e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-x*log(x^2)-20*exp(x)*x-4*x+2)/exp(x)/x,x, algorithm="maxima")

[Out]

1/4*e^(-x)*log(x^2) - 5*x + e^(-x)

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Fricas [A]
time = 0.34, size = 19, normalized size = 0.76 \begin {gather*} -\frac {1}{4} \, {\left (20 \, x e^{x} - \log \left (x^{2}\right ) - 4\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-x*log(x^2)-20*exp(x)*x-4*x+2)/exp(x)/x,x, algorithm="fricas")

[Out]

-1/4*(20*x*e^x - log(x^2) - 4)*e^(-x)

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Sympy [A]
time = 0.09, size = 14, normalized size = 0.56 \begin {gather*} - 5 x + \frac {\left (\log {\left (x^{2} \right )} + 4\right ) e^{- x}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-x*ln(x**2)-20*exp(x)*x-4*x+2)/exp(x)/x,x)

[Out]

-5*x + (log(x**2) + 4)*exp(-x)/4

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Giac [A]
time = 0.39, size = 18, normalized size = 0.72 \begin {gather*} \frac {1}{4} \, e^{\left (-x\right )} \log \left (x^{2}\right ) - 5 \, x + e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-x*log(x^2)-20*exp(x)*x-4*x+2)/exp(x)/x,x, algorithm="giac")

[Out]

1/4*e^(-x)*log(x^2) - 5*x + e^(-x)

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Mupad [B]
time = 1.50, size = 18, normalized size = 0.72 \begin {gather*} {\mathrm {e}}^{-x}-5\,x+\frac {\ln \left (x^2\right )\,{\mathrm {e}}^{-x}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(x + (x*log(x^2))/4 + 5*x*exp(x) - 1/2))/x,x)

[Out]

exp(-x) - 5*x + (log(x^2)*exp(-x))/4

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